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Given: $$p = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&5 &7 &4 &6 &9 &3 &2 &8 \end{bmatrix}$$ How would I go about determining $k$ such that: $$p^k = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&2 &3 &4 &5 &6 &7 &8 &9 \end{bmatrix}$$

I know I can continuously multiply the permutations till I reach the answer but is there a quicker or "right way" of determining the answer?

Damian Jacobs
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    Writing $p$ in its disjoint cyclic representation, this is $(2~5~6~9~8)(3~7)$. Look at the least common multiple of the cycle's lengths in the disjoint cyclic representation. – JMoravitz Sep 29 '20 at 20:43
  • @JMoravitz I've added onto my question, if you wouldn't mind having a look. – Damian Jacobs Sep 29 '20 at 21:12
  • You must have made some mistake in your example for $p_2$. The image of $1$ under $(p_2)^4$ is $6$. That should be quick to verify since the image of $1$ under $(p_2)^k$ is simply going to be the $k$'th term in the sequence (starting count from zero) of $1\mapsto 5\mapsto 3\mapsto 2\mapsto 6\mapsto 4\mapsto 7\mapsto 8\mapsto 1\mapsto \cdots$, the first positive $k$ for which $1$ maps to itself (and indeed all elements map to themselves for your example of $p_2$) being $8$ – JMoravitz Sep 29 '20 at 21:13
  • See here and here and your textbook and/or lecture notes, and many other sources. – JMoravitz Sep 29 '20 at 21:20
  • @JMoravitz I did make a mistake indeed. I was multiplying $p^2$ by $p^2$ and then $p^4$ by $p^4$. Apologies – Damian Jacobs Sep 29 '20 at 21:21

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